Obtain an expression for the frequency of radiation emitted when a hydrogen atom de-excites from level n to level (n−1). F
Question:

Obtain an expression for the frequency of radiation emitted when a hydrogen atom de-excites from level to level (n−1). For large n, show that this frequency equals the classical frequency of revolution of the electron in the orbit.

Solution:

It is given that a hydrogen atom de-excites from an upper level (n) to a lower level (n−1).

We have the relation for energy (E1) of radiation at level as:

$E_{1}=h v_{1}=\frac{h m e^{4}}{(4 \pi)^{3} \in_{0}^{2}\left(\frac{h}{2 \pi}\right)^{3}} \times\left(\frac{1}{n^{2}}\right)$    …(1)

Where,

$v_{1}=$ Frequency of radiation at level $n$

$h=$ Planck’s constant

$m=$ Mass of hydrogen atom

$e=$ Charge on an electron

$\epsilon_{0}=$ Permittivity of free space

$E_{2}=h v_{2}=\frac{h m e^{4}}{(4 \pi)^{3} \in_{0}^{2}\left(\frac{h}{2 \pi}\right)^{3}} \times \frac{1}{(n-1)^{2}}$    …(2)

Now, the relation for energy $\left(E_{2}\right)$ of radiation at level $(n-1)$ is givenas:

Where,

$\mathrm{v}_{2}=$ Frequency of radiation at level $(n-1)$

Energy (E) released as a result of de-excitation:

E = E2E1

hν = E2 − E1 … (iii)

Where,

ν = Frequency of radiation emitted

Putting values from equations (i) and (ii) in equation (iii), we get:

$v=\frac{m e^{4}}{(4 \pi)^{3} \epsilon_{0}^{2}\left(\frac{h}{2 \pi}\right)^{3}}\left[\frac{1}{(n-1)^{2}}-\frac{1}{n^{2}}\right]$

$=\frac{m e^{4}(2 n-1)}{(4 \pi)^{3} \in_{0}^{2}\left(\frac{h}{2 \pi}\right)^{3} n^{2}(n-1)^{2}}$

For large $n$, we can write $(2 n-1) \simeq 2 n$ and $(n-1) \simeq n$.

$\therefore v=\frac{m e^{4}}{32 \pi^{3} \in_{0}^{2}\left(\frac{h}{2 \pi}\right)^{3} n^{3}}$     …(iv)

Classical relation of frequency of revolution of an electron is given as: $v_{\mathrm{c}}=\frac{v}{2 \pi r}$    …(v)

Where,

Velocity of the electron in the nth orbit is given as:

$v=\frac{e^{2}}{4 \pi \in_{0}\left(\frac{h}{2 \pi}\right) n}$    …(vi)

And, radius of the nth orbit is given as:

$r=\frac{4 \pi \in_{0}\left(\frac{h}{2 \pi}\right)^{2}}{m e^{2}} n^{2}$    …(vii)

Putting the values of equations (vi) and (vii) in equation (v), we get:

$v_{c}=\frac{m e^{4}}{32 \pi^{3} \in_{0}^{2}\left(\frac{h}{2 \pi}\right)^{3} n^{3}}$    …(viii)

Hence, the frequency of radiation emitted by the hydrogen atom is equal to its classical orbital frequency.

 

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